# Prove a square can't be written $5x+ 3$, for all integers $x$.

Homework question, should I use induction?.. Help please

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what you tried? –  Felipe Jan 30 at 16:07

What are the only possibilities for the last digit of $5x + 3$ ?

Squares can only have as last digit: $0 , \ 1 , \ 4 , \ 5 , \ 6 , \ 9$.

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I have up voted this but do not know if this is suited as an answer... i am sure it would get more up votes if it is written as a hint by saying something like... "what are the only possibilities of last digits of $5x+3$" this is good enough though :) –  Praphulla Koushik Jan 30 at 16:16
Congrats for the nice answer badge, Neo. –  Balarka Sen Jan 31 at 12:53

Hint $\$ Let $\,n\,$ by any integer. $\ {\rm mod}\ 5\!:\ n \equiv 0,\pm1,\pm2\,\Rightarrow\, n^2\equiv 0,1,4,\,$ so $\ n^2\not\equiv 3\!\pmod{\!5}$

Remark  It's easier if you know $\mu$Fermat: $\ 3 \equiv n^2\overset{\rm square}\Rightarrow\color{#0a0}{-1}\equiv \,n^4\ [\,\equiv \color{#c00}1$ by $\mu$Fermat]

This is a special case of Euler's Criterion: $\,{\rm mod}\ p=5\!:\,\ 3 \not\equiv n^2\,$ by $\ 3^{(p-1)/2}\equiv\color{#0a0}{-1}\not\equiv \color{#c00}1$

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Consider the squares under $\mod{10}$ $$1^2\equiv{1}\pmod{10}$$ $$2^2\equiv{4}\pmod{10}$$ $$3^2\equiv{9}\pmod{10}$$ $$4^2\equiv{6}\pmod{10}$$ $$5^2\equiv{5}\pmod{10}$$ $$...$$ $$9^2\equiv{1}\pmod{10}$$ You will then see that numbers that end in 3 can not written as squares, since the only numbers that can end squares are $0,1,4,5,6,9$...

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Let $A=\{y\in\mathbb{Z}: y=5x+3, x\in\mathbb{R} \}$

And, $y\in A \Rightarrow y$ are a multiple of 5 plus 3! The set of numbers that are multiples of 5 are $\{...,-10,-5,0,5,10,...\} = \{n5: n \mathbb{Z} \}$